MIT differential equations 19,20,21,22

laplace transform

laplace transform comes from power series.

sigma(0->inf) an x^n = A(x)

this is a discrete but write it as

sigma(0->inf) a(n) x^n = A(x) 

suppose  a(n) = 1, A(x) = 1/(1-x) for x in (-1,1)

suppose  a(n) = 1/n!, A(x) = e^x

continuous analog 

$(0->inf) a(t) x^t dt = A(x)

let x^t = e^(t Inx)

let -s = lnx 

Then, we can get $(0->inf) f(t) e^-st dt = F(s) which is laplace transform

This may seem a little bit wired as x is mapping to s.

Notice the difference between operator and transform.

Think of the black box analog.

f(t) ->  TRANSFORM -> F(s)

f(t) ->  OPERATOR    -> g(s)

we'll do a lot of improper integral.  $(0->inf) f(t) dt = lim(R=0->inf)  $(0->R)f(t)dt

Linearity of laplace transform

L(f + g) = L(f) + L(g)

f(t) -------------->   F(s)   (s>0)

1                           |  1/s 

e^at f(t)               |    F(s-a)   s > a      exponential shift law

  e^at                  |         1/(s-a)             

 e^a+bi t             |          1/(s-a-bi)

cos at = (e^iat + e^-iat)/2 | 1/(s-ia) + 1/(s+ia)   /2

sin at = (e^iat - e^-iat)/2  |   1/(s-ia) - 1/(s-ia) /2

Inverse Laplace Transform

convert F(s) back to f(t) that you are looking for

The technique involves partial factor decomposition 

1/s(s+3)  =  1/3   1/s - 1/(s+3)

= 1/3  ( 1 - e^-3t)

L(t^n) = n!/s^(n+1)                

Lesson 20

notice Laplace Transform is an improper integral, which means it doesn't converge all the time.

We can f(t) an exponential type when  | f(t) | <= ce^kt

|sinx| is an exponential type for c = 1; k = 0;

| t^n| is an exponential type because  t^n / e^kt  converges as t goes to INF.

However | 1/x |,  | e^(t^2) | isn't an exponential type.

L.T. to solve second order ODE. It must be an IVP problem.  yo, y'o

f'(t) = sF(s) - f(0)

f''(t) = L([f'(t)]') = s( F(f'(t) ) - f'(0)

Lesson 21

Convolution

 f(t) * g(t) := $(0->t) f(t-u) g(u) du

What' the L.T of f(t) g(t)?

Write out the defination of L(f(t) and L(g(t)). You can easily get the formula.

In the analog of power series, sigma(An x^n), sigma(Bn x^n), Let Cn be the coefficient of the product of these two series. 

Application:

Radioactive dumping. 

Suppose there's a nuclear plant which keeps dump nuclear wastes to a pile at rate of f(t), then in t1, it will dump  t1 *  f(t) wastes.

But the waste will decay at the same time at e^-kt.

Then the overall nuclear wastes will be $(0->t) f(u) e^-k(t-u) du

Lesson 22

L.T ODE jumping continuity 

FIrst, introduce "unit step" function. u(t):= 0 for t < 0; 1 for t > 0;  undef for t = 0;

Uab(t):= Ua(t) - Ub(t) = U(t-a) - U(t-b)

f(t) * Uab(t) = f(t) for a < t < b; 0 elsewhere

Now consider L(U(t)) and L(1). Both of them are equal to 1/s. But what's the inverse L.T of 1/s?

Here comes the uniqueness problem of L.T. 

L.T is an improper integral from 0 to INF, therefore, it only cares about the future. That's why many engineers and scientists use L.T to care about the future, not the past.

For the uniqueness of L.T., we define F(s) ---L-1-----> U(t)f(t)

Question: Is there a equation to map L(f(t)) to L(f(t-a))? Sadly, there's no such equation because L.T will ignore a fraction of f(t) where t < 0. 

But there's a equation to do similar thing but you have to write it in a particular way.

U(t-a) f(t-a) -----L-------> e^-as F(s) = e^-as L[f(t+a)]

Notice the exponent will tell you the point of discontinuity;

EX: L(Uab(t)) = L(U(t-a) - U(t-b)) = 1/s (e^-as - e^-bs)

L(U(t-1) t^2)  = e^-s L[(t+1)^2] .....

L^-1 [  (1+e^-PI s) / ( s^2 + 1) ]  = L^-1 [  1/(s^2+1)  + e^-PI s / (s^2 + 1)   ]  =  u(x) sinx + u(t-PI) sin(t - PI)

Make cases: 

x in [0, PI]  L^-1 = sint

x in [PI, INF]  L^-1 = 0;